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Question Number 187050 Answers: 0 Comments: 1
$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\underset{\mathrm{0}} {\int}^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \:\:\frac{\boldsymbol{{cos}}^{−\mathrm{1}} \boldsymbol{{x}}\:+\:\boldsymbol{{cos}}\:\boldsymbol{{x}}}{\boldsymbol{{Ln}}\:\boldsymbol{{x}}}\:\:\boldsymbol{{dx}}\:=??\:\:\: \\ $$$$ \\ $$
Question Number 187046 Answers: 1 Comments: 0
$${find}\:\:\:{the}\:{general}\:\:{solution}\:{for}\:\:{the}\:{following} \\ $$$${system}\:{of}\:{equations} \\ $$$$\left(\frac{{dx}_{\mathrm{1}} }{{dt}}\right)=\mathrm{2}{x}_{\mathrm{1}} +\mathrm{2}{x}_{\mathrm{2}} \\ $$$$\left(\frac{{dx}_{\mathrm{2}} }{{dt}}\right)={x}_{\mathrm{1}} +\mathrm{3}{x}_{\mathrm{2}} \\ $$$$ \\ $$
Question Number 187029 Answers: 1 Comments: 0
Question Number 187027 Answers: 1 Comments: 0
Question Number 187025 Answers: 1 Comments: 0
Question Number 187044 Answers: 0 Comments: 0
$${f}\left({x},{y}\right)=\left\{\left({x}+{y}\right),\mathrm{0}<{x}<\mathrm{1},\:\mathrm{0}<{y}<\mathrm{1}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0},\:{otherwise} \\ $$$${find}: \\ $$$$\left(\mathrm{1}\right)\:\sigma{xy} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\:{p}\left(\mathrm{0}<{x}<\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\mid\left(\frac{\mathrm{2}}{\mathrm{5}}\right)>{y}>\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right) \\ $$$$ \\ $$$${note}:\:{f}\left({x},{y}\right)\:{is}\:{the}\:{joint}\:{PDF} \\ $$
Question Number 187020 Answers: 0 Comments: 1
$$\frac{{a}}{{x}}=\frac{{b}}{{y}}=\frac{{c}}{{z}}=\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:,{a}−\mathrm{2}{b}+{c}=\mathrm{2}\:\:{and}\:\:\mathrm{2}{y}−\mathrm{3}{z}=\mathrm{1}\:\:\:\:{x}=? \\ $$$${how}\:{is}\:{solution} \\ $$
Question Number 187011 Answers: 0 Comments: 0
$$\mathrm{a}\in\mathbb{C} \\ $$$$\begin{vmatrix}{\mathrm{a}\:\:\:\mathrm{1}\:\:\:\:\ldots\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\ddots\:\:\ddots\:\:\vdots}\\{\vdots\:\:\ddots\:\:\ddots\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\ldots\:\:\:\:\mathrm{1}\:\:\:\mathrm{a}}\end{vmatrix}=? \\ $$
Question Number 187010 Answers: 0 Comments: 0
$$\mathrm{a},\mathrm{b}\in\mathbb{C} \\ $$$$\:\begin{vmatrix}{\mathrm{a}+\mathrm{b}\:\:\:\:\:\:\:\:\:\mathrm{a}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\ldots\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\:\:\:\mathrm{b}\:\:\:\:\:\:\:\:\:\:\mathrm{a}+\mathrm{b}\:\:\:\:\:\:\ddots\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\:\:\:\:\vdots}\\{\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\:\:\vdots\:\:\:\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}}\\{\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\ldots\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}+\mathrm{b}}\end{vmatrix}=? \\ $$
Question Number 187008 Answers: 2 Comments: 0
$$\frac{{a}}{\mathrm{3}}=\frac{{b}}{\mathrm{4}}=\frac{{c}}{\mathrm{5}}\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}{a}+{c}=\mathrm{42}\:\:\:\:\:\:\:\:{b}=? \\ $$$${how}\:{is}\:{solution} \\ $$
Question Number 187007 Answers: 0 Comments: 2
$${prove}\:\:{that} \\ $$$$\mathrm{7}^{{k}+\mathrm{1}} \mid\:\mathrm{13}^{\mathrm{7}{k}} \:+\:\mathrm{1}\:,\:\forall{k}\in\mathbb{N} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Question Number 186999 Answers: 0 Comments: 0
Question Number 186998 Answers: 1 Comments: 1
Question Number 186996 Answers: 1 Comments: 0
Question Number 186989 Answers: 0 Comments: 0
Question Number 186983 Answers: 3 Comments: 0
Question Number 186960 Answers: 1 Comments: 0
Question Number 186958 Answers: 0 Comments: 0
Question Number 186952 Answers: 2 Comments: 0
Question Number 186951 Answers: 2 Comments: 0
Question Number 186950 Answers: 0 Comments: 0
Question Number 186941 Answers: 0 Comments: 1
$$\frac{{sinx}}{{x}^{{b}} }=\frac{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{x}^{\mathrm{2}{n}+\mathrm{1}} }{{x}^{{b}} } \\ $$$${prove}\:{that} \\ $$
Question Number 186929 Answers: 3 Comments: 0
$$\:\:{If}\:\begin{cases}{{B}+{R}+{P}=−\mathrm{1}}\\{{B}^{\mathrm{2}} +{R}^{\mathrm{2}} +{P}^{\mathrm{2}} =\mathrm{17}}\\{{B}^{\mathrm{3}} +{R}^{\mathrm{3}} +{P}^{\mathrm{3}} =\mathrm{11}}\end{cases} \\ $$$$\:{then}\:{B}^{\mathrm{5}} +{R}^{\mathrm{5}} +{P}^{\mathrm{5}} \:=? \\ $$
Question Number 186926 Answers: 1 Comments: 1
Question Number 186924 Answers: 3 Comments: 0
Question Number 186923 Answers: 0 Comments: 0
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